Prove that the line joining midpoints of two sides is parallel to the third

Question

In triangle ABC, D is the midpoint of AB and E is the midpoint of AC. Prove that DE is parallel to BC and DE = BC/2. (This is the Midpoint Theorem.)

Solution — Step by Step

Given: In △ABC, D is the midpoint of AB and E is the midpoint of AC. So $AD = DB = \frac{1}{2}AB$ and $AE = EC = \frac{1}{2}AC$ .

To prove: (i) DE ∥ BC, and (ii) DE = $\frac{1}{2}$ BC.

Since D is the midpoint of AB: $\frac{AD}{AB} = \frac{1}{2}$ , which gives $\frac{AD}{DB} = 1$ .

Since E is the midpoint of AC: $\frac{AE}{AC} = \frac{1}{2}$ , which gives $\frac{AE}{EC} = 1$ .

Therefore: $\frac{AD}{DB} = \frac{AE}{EC} = 1$

This is the condition of the Basic Proportionality Theorem (BPT, also called Thales’ theorem) in reverse — the converse of BPT.

Converse of BPT: If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.

Since $\frac{AD}{DB} = \frac{AE}{EC}$ , by the converse of BPT:

\mathbf{DE \parallel BC} \quad \text{(proved)}

Since DE ∥ BC and △ADE is formed within △ABC with DE cutting through D and E:

In △ADE and △ABC:

$\angle A = \angle A$ (common angle)
$\angle ADE = \angle ABC$ (corresponding angles, since DE ∥ BC)
Therefore △ADE ~ △ABC (by AA similarity)

From the similarity:

\frac{DE}{BC} = \frac{AD}{AB} = \frac{1}{2}

\mathbf{DE = \frac{1}{2}BC} \quad \text{(proved)}

Why This Works

The Midpoint Theorem is really just the Basic Proportionality Theorem in a special case. When the ratio is exactly $1 : 1$ on both sides, the line connecting the division points is always parallel to the base and exactly half as long. The similarity ratio between the small triangle and the large one is $1 : 2$ , so all corresponding lengths are in the same ratio.

The Midpoint Theorem has a useful converse: if a line is drawn through the midpoint of one side of a triangle, parallel to the second side, it bisects the third side. This is used frequently in proofs about quadrilaterals (e.g., the diagonals of a parallelogram bisect each other).

Alternative Method

Using coordinate geometry:

Let $A = (0, 0)$ , $B = (2a, 0)$ , $C = (2b, 2c)$ .

Midpoint D of AB: $D = (a, 0)$

Midpoint E of AC: $E = (b, c)$

Slope of DE: $\frac{c - 0}{b - a} = \frac{c}{b-a}$

Slope of BC: $\frac{2c - 0}{2b - 2a} = \frac{2c}{2(b-a)} = \frac{c}{b-a}$

Equal slopes → DE ∥ BC ✓

$DE = \sqrt{(b-a)^2 + c^2}$ , $BC = \sqrt{(2b-2a)^2 + (2c)^2} = 2\sqrt{(b-a)^2 + c^2} = 2 \cdot DE$ ✓

Common Mistake

Students often try to prove this using congruence (SSS, SAS, etc.) rather than similarity, and get stuck because DE and BC have different lengths. The correct tool is similarity — two triangles with the same angles but different sizes. Switch from congruence to similarity whenever you see “half the length” or a scale factor other than 1.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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